Guest Posted December 13, 2009 Report Share Posted December 13, 2009 (edited) 0, 2, 8, 18, 7, 7, 9, 13, 10, 9, 1, 2, 5, . . . If you can figure out the pattern, then let A=1, B=2, C=3, . . . and find the first place in the sequence where nine consecutive letters spell out two consecutive English words (letters are in order). Without giving away the puzzle, include synonyms of both words somewhere in your response. Edited December 13, 2009 by xamdam Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 13, 2009 Report Share Posted December 13, 2009 Any hints? Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 13, 2009 Report Share Posted December 13, 2009 I found a simple-to-generate cyclic 7920 pattern that contains this sequence but there are no words in it as you describe. In my generation scheme, the next 3 terms (after the 5) are 6, 26, 17. Perhaps you could supply more terms. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 13, 2009 Report Share Posted December 13, 2009 Reveal hidden contents My function generates an infinite, non-cyclical sequence. The first 21 elements are : 0, 2, 8, 18, 7, 7, 9, 13, 10, 9, 1, 2, 5, 10, 7, 7, 8, 11, 7, 4, 4, … Both the minimum and maximum occur above. While there are infinite occurrences of the maximum, the minimum occurs only once. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 14, 2009 Report Share Posted December 14, 2009 Reveal hidden contents You might be able to find the key to the basic pattern by focusing on the first four elements. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 14, 2009 Report Share Posted December 14, 2009 (edited) On 12/14/2009 at 6:18 PM, xamdam said: Reveal hidden contents You might be able to find the key to the basic pattern by focusing on the first four elements. I was wondering earlier whether the first one or two or so elements were arbitrarily set, or if they were determined by the function. Like, in the Fibonacci sequence (0, 1, 1, 2, 3, 5, 8...), the first two numbers are essentially random. So I guess this answers that question My biggest question would be this: is it a piecewise function (like, it undergoes a different equation if xn > 9 than does xn < 10), or is it the same kinda thing everywhere along the sequence? Edited December 14, 2009 by DarthNoob Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 15, 2009 Report Share Posted December 15, 2009 (edited) The closest thing I can get is this: Reveal hidden contents if an < 10 a(n+1) = 2a(n) + 2 - a(n-1) and obviously something to handle cases where the number would go negative But it doesn't work everywhere Edited December 15, 2009 by DarthNoob Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 15, 2009 Report Share Posted December 15, 2009 On 12/14/2009 at 11:11 PM, DarthNoob said: I was wondering earlier whether the first one or two or so elements were arbitrarily set, or if they were determined by the function. Like, in the Fibonacci sequence (0, 1, 1, 2, 3, 5, 8...), the first two numbers are essentially random. So I guess this answers that question My biggest question would be this: is it a piecewise function (like, it undergoes a different equation if xn > 9 than does xn < 10), or is it the same kinda thing everywhere along the sequence? Reveal hidden contents Unlike the first two elements of the Fibonacci sequence, all elements are determined by the function. It is not a piecewise function -- the sequence can be generated with one simple spreadsheet formula (containing no "if" statements or conditions on the domain). However, the way the function behaves, it can appear to be in two (or even three) pieces: 0, 2, 8, 18, then 7, 7, 9, 13, 10, 9, then behaving nicely from here on 1, 2, 5, 10, 7, 7, 8, 11, 7, 4, 4, … Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 15, 2009 Report Share Posted December 15, 2009 On 12/15/2009 at 12:31 AM, DarthNoob said: The closest thing I can get is this: Reveal hidden contents if an < 10 a(n+1) = 2a(n) + 2 - a(n-1) and obviously something to handle cases where the number would go negative But it doesn't work everywhere Reveal hidden contents If I'm understanding your equation right, a(n+1) = 2a(n) - a(n-1) it does predict the 7th, 8th, 10th, 13th and 14th terms based on the previous term. However, since my range ( or codomain?), unlike the Fibonacci sequence, is a finite set of integers (see previous hint re minimum and maximum) and the sequence is infinite I don't think it would be possible to write a formula that looks at one or two terms and predicts the next, would it? It seems like any finite sub-sequence you picked would eventually have to appear again and again in this infinite string. How about if I tell you that this sequence can be generated if the domain of the function is the sequence of whole numbers (non-negative integers). (There is another domain that happens to work too.) The rule for the function is probably simplest to say in words (about 7 words), or to write as a spreadsheet formula. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 15, 2009 Report Share Posted December 15, 2009 Reveal hidden contents "It seems like any finite sub-sequence you picked would eventually have to appear again and again in this infinite string. " My bad. I was thinking of random numbers. You could easily design a sequence where certain sub-sequences don't repeat: 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, ... (here, 0, 1, 1, 0 won't repeat). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 15, 2009 Report Share Posted December 15, 2009 Guess I'll be the whistle-blower and say this one's not solvable; you couldn't crack it open with a machete. Thanks though, Xamdam, for wasting my time! Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 15, 2009 Report Share Posted December 15, 2009 (edited) Reveal hidden contents each member of the sequence is the sum of the most significant digit of some number (which is 0 only at 0) with another digit from the same number. This number may be a simple function of a positive integer, but there are sooooo many of them! Edited December 15, 2009 by superprismatic Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 15, 2009 Report Share Posted December 15, 2009 Reveal hidden contents does this have to do with the mod operator, for ex 512 mod 16 =0 512 mod 17 = 2 512 mod 18= 8 512 mod 19 = 18 However this function does not predict the numbers correctly after the first 4. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2009 Report Share Posted December 16, 2009 On 12/15/2009 at 5:27 PM, ljb said: Guess I'll be the whistle-blower and say this one's not solvable; you couldn't crack it open with a machete. Thanks though, Xamdam, for wasting my time! And here I am working on your Twisted Decimal Code, 12 hours a day, staying home from work, 8 computers cranking away full-time, for a few weeks now, and you quit in three days?! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2009 Report Share Posted December 16, 2009 On 12/15/2009 at 8:24 PM, superprismatic said: Reveal hidden contents each member of the sequence is the sum of the most significant digit of some number (which is 0 only at 0) with another digit from the same number. This number may be a simple function of a positive integer, but there are sooooo many of them! Reveal hidden contents Well done. You're on the mark so far. Most kids probably learn the fuction around 4th to 6th grade. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2009 Report Share Posted December 16, 2009 (edited) On 12/15/2009 at 8:39 PM, vimil said: Reveal hidden contents does this have to do with the mod operator, for ex 512 mod 16 =0 512 mod 17 = 2 512 mod 18= 8 512 mod 19 = 18 However this function does not predict the numbers correctly after the first 4. Reveal hidden contents No, it's a bit simpler than that. That's a cool idea for your next puzzle post though. (Check out superprismatic's reply and my response.) Edited December 16, 2009 by xamdam Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2009 Report Share Posted December 16, 2009 Reveal hidden contents My function generates an infinite, non-cyclical sequence. The first 21 elements are : 0, 2, 8, 18, 7, 7, 9, 13, 10, 9, 1, 2, 5, 10, 7, 7, 8, 11, 7, 4, 4, … Both the minimum and maximum occur above. While there are infinite occurrences of the maximum, the minimum occurs only once. Reveal hidden contents 5, 8, 14? Maybe it has to do with being right brained or left brained. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2009 Report Share Posted December 16, 2009 (edited) On 12/16/2009 at 4:43 AM, Patouie said: Reveal hidden contents 5, 8, 14? Maybe it has to do with being right brained or left brained. Reveal hidden contents Firstly, those are indeed the 21st, 22nd and 23rd terms. Lucky guess? And lastly, good luck finding the words. Are you using a spreadsheet? Edited December 16, 2009 by xamdam Quote Link to comment Share on other sites More sharing options...
0 DudleyDude Posted December 16, 2009 Report Share Posted December 16, 2009 I have to agree with ljb. Xamdam must be some kind of rat to try to stick us with this problem. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2009 Report Share Posted December 16, 2009 (edited) On 12/16/2009 at 5:45 AM, xamdam said: Reveal hidden contents Firstly, those are indeed the 21st, 22nd and 23rd terms. Lucky guess? And lastly, good luck finding the words. Are you using a spreadsheet? Reveal hidden contents I don't have a spreadsheet at home. I'll try that from work -- I'll definitely need one. (And no, it wasn't a lucky guess, you cruel person!) Thinking up a new use for LJB's machete... Edited December 16, 2009 by Patouie Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2009 Report Share Posted December 16, 2009 On 12/16/2009 at 3:33 PM, DudleyDude said: I have to agree with ljb. Xamdam must be some kind of rat to try to stick us with this problem. . . . . Whata buncha lightweights Quote Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2009 Report Share Posted December 16, 2009 I caught a glimpse of W's brother with some hot chicks, but in the end, I couldn't cut the mustard. Xamdam is definitely a weasel. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 17, 2009 Report Share Posted December 17, 2009 (edited) Kudos to ljb, DudleyDude and Patouie. Superprismatic gets credit for an assist. Very nice. Edited December 17, 2009 by xamdam Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 17, 2009 Report Share Posted December 17, 2009 Reveal hidden contents Several questions unarg e-------------------------------u c-------------------------------a n-------------------------------b e--------------------------------l u-------------------------------y q-------------------------------- e-------------------------------a s-----hcae gnidrager esir Quote Link to comment Share on other sites More sharing options...
0 plainglazed Posted December 17, 2009 Report Share Posted December 17, 2009 I confess, cant cut it anymore, either. Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 17, 2009 Report Share Posted December 17, 2009 Well, just call me 'Saber Skunk' from now on 'cause I misread the clue in post #9 to say 'positive' instead of 'non-negative' which really cut me to the quick! Quote Link to comment Share on other sites More sharing options...
Question
Guest
0, 2, 8, 18, 7, 7, 9, 13, 10, 9, 1, 2, 5, . . .
If you can figure out the pattern, then let A=1, B=2, C=3, . . .
and find the first place in the sequence where nine consecutive letters
spell out two consecutive English words (letters are in order).
Without giving away the puzzle, include synonyms of both words somewhere in your response.
Edited by xamdamLink to comment
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